Day_Lecture_7

Day_Lecture_7 - Chemistry 307 Chapter 7 When you learned...

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1 Chemistry 307 Chapter 7 When you learned about the S N 2 reaction, you had considered two possible mechanisms, involving either a single step (nucleophilic substitution with second order kinetics, S N 2) or two steps, i.e., 1. R–LG R + + LG 2. R + + Nu R–Nu The reactions you learned about so far followed the one- step mechanism; you learned about the kinetic order and about the role of nucleophile, leaving group, the structure of the substrate, and of the solvent. These reactions proceeds via transition states.
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2 The second mechanism that you considered was a two- step reaction. In the first step, the haloalkane undergoes ionic dissociation forming a carbocation, followed in the second step by reaction with a nucleophile to complete the substitution. This mechanism would have first- order kinetics: the relative slow formation of the carbocation, R + (step 1), would determine the rate (k [R–LG]); once the cation is generated, it would capture the nucleophile rapidly, forming the product. Does the second type of nucleophilic substitution, involving two steps, actually occur? To answer this question we compare two reactions with two different substrates: a) the reaction of iodide ion (Na + I ) with bromoalkanes in acetone (Table 6.8); bromomethane reacts much faster with Na + I than 2-bromo-2-methylpropane; this result is exactly what you learned to expect from an S N 2 reaction; b) the reaction of water, H 2 O, with the same bromoalkanes (Table 7.1); this reaction proceeds much faster for 2-bromo-2-methylpropane than for bromomethane; the reversed order of reactivity is
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3 incompatible with an S N 2 reaction and requires a different mechanism. A more detailed study shows that the reaction with water follows a first order rate, as expected for the two-step reaction. The reaction also has different stereochemical consequences: the stereochemistry at the reacting carbon is randomized, i.e., iodide ion converts enantiomerically pure 3-bromo-3- methylhexane to racemic 3-iodo-3-methylhexane. This result requires an achiral intermediate, identified as the planar
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Day_Lecture_7 - Chemistry 307 Chapter 7 When you learned...

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